FINITELY GENERATED CLASSES OF SETS OF NATURAL NUMBERS

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1 FINITELY GENERATED CLASSES OF SETS OF NATURAL NUMBERS JULIA ROBINSON We say that a set S of natural numbers is generated by a class S of functions if S is the range of a function obtained by composition from the functions of ï. Here we consider the identity function / to be obtained by the empty composition. We consider only the case in which 5e consists of functions of one variable on and to the set of natural numbers N. All sets considered are subsets of N. A class C of sets is generated by?f if every nonempty set of 6 is generated by SF and every set generated by SF is in Q. Q is finitely generated if there is a finite class of functions SF which generates C. A function is compatible with C if F(S) belongs to 6 for every Sin e. Clearly, if SF generates C then every in ÍF is compatible with C. Also N belongs to C. Furthermore to show that!f generates C it is sufficient to prove that every GSF is compatible with C, NE C, and every nonempty set in C is generated by ÍF. In [l] finite sets of relatively simple functions are given which generate the classes of recursively enumerable sets and diophantine sets. Here we ask: What classes of sets can be finitely generated? In particular, we will show that if G is a denumerable field of sets such that all finite sets belong to C and at least one infinite set having an infinite complement belongs to C, then Q is finitely generated. Lemma 1. If 3 is an infinite class of infinite sets then there is a set M such that MC\ T and MC\ T are infinite for every T belonging to 3. Proof. Let each T in 3 be listed infinitely often in the sequence To, Ti,. We will construct M and M simultaneously. At the «th step put the least unassigned number of Tn in M and then the least remaining unassigned number of Tn in M. Lemma 2. Given infinite sets To, Ti, > and nonempty sets So, Si, such that Sk = S whenever Tkr\Tt is not empty. Then there is a function F such that F(Tn) =S. Proof. Make a list of natural numbers such that each number occurs infinitely often. We will use the list to define at one argument for each step of the construction. Namely, if at a given step n is the Received by the editors April 29,

2 FINITELY GENERATED CLASSES 609 next number in the list and if x is the least unassigned number in Tn and y is the least number in S not yet in F(Tn), put Fx = y; while if there is no such y, put Fx = the least number of Sn- Finally, in case UTk^N, put Fx = 0 for every x not in any 7*. Theorem 1. Hypotheses. (1) 6 is a denumerable class of sets. (2) If A and B belong to 6 then AVJB belongs to 6. (3) N belongs to Q. (A) Either (i) C contains all finite sets or (ii) C contains only infinite sets and 6 contains the set obtained from a member set by adjoining a single element. (5) There is an infinite partition (PC 6 of N into infinite sets all of which are generated by a finite class 5 of functions compatible with e. Conclusion, Q is finitely generated. Proof. Let S>= {P0, Pi, }. We say that I is the index set of a set S if 7 = {n : SC\ Pn9i0). Let M be the set given by Lemma 1 such that every infinite index set of a set belonging to C contains infinitely many elements of both M and M. (Concerning M, we shall use only that M is infinite.) List the nonempty sets of C in a sequence T0, Ti, so that Tm = N for every m in M. We will define F in such a way that (a) F(Tnr\Pk) = Tk whenever Tnr\Pk is infinite, and (b) F(Tn) =N whenever Tn has an infinite index set. Let us check that this is sufficient. In the first place, every nonempty set of Q will be generated by {F} UEF since F(Pk) = Tk. Also F is compatible with 6, i.e. for every «, F(Tn) belongs to C. To see this, let 7 be the index set of TV If I is infinite then F(Tn) = N and N belongs to C by (3). If I is finite, then F(Tn) = U F(Tnr\Pk), kei where F(Tni^Pk) is either finite or Tk according as TnC\Pk is finite or infinite. Hence if C contains all finite sets then F(Tn) is a nonempty finite union of sets of Q and so F(Tn) belongs to C. If C consists only of infinite sets, then TnC\Pk is infinite for some k. Thus F(Tn) is the union of a nonempty finite union of sets of C and a finite set. Hence F(Tn) belongs to Q. Therefore to complete the proof, we need only define F satisfying (a) and (b). Now make a list of pairs («, k) so that each pair occurs infinitely often. We will use this list to index the steps in the definition of F. At a step corresponding to (n, k), if there is a number in TnP\Pk at which F has not yet been assigned, let the least such number be x. Put Fx equal to the least number in Tk which is not already in F(TnC\Pk) if there is such a number; otherwise put Fx equal to the

3 610 JULIA ROBINSON [June least number in Tk. Also if the index set of Tn is infinite and y is the least number unassigned in TnC\Pm for any m in M, define Fy=k. Then will satisfy both (a) and (b). Examples. The classes of recursive sets, recursively enumerable sets, arithmetical sets, hyperarithmetical sets, are all finitely generated. We can let Pk = (SD)kD(N) since S, D are compatible with each of these classes. Here 5 is the successor function and D the double function. By the usual diagonal argument it is easy to see that the class of arithmetical sets cannot be generated by a finite class of arithmetical functions. (Similarly hyperarithmetical sets cannot be generated by a finite class of hyperarithmetical functions.) I do not know if the class of recursive sets can be generated by a finite number of recursive functions. An interesting class of sets which is not covered by Theorem 1 is the class of polynomial sets and I do not know whether it is finitely generated. By a polynomial set, I mean the range of a polynomial with positive integer coefficients and any number of variables. If we assume that C is a field of sets which contains all finite sets, then condition (5) in Theorem 1 can be weakened to the mere existence of (P. In order to prove this, we shall need the following combinatorial result. Lemma 3. Let Qbe a denumer able field of sets which contains all finite sets. Suppose further that there is an infinite partition (P0 of N into infinite sets of Q. Then there is an infinite partition (P of N into infinite sets Pa, Px, of C such that for every S in Q either S or S is included in a finite union of the P's. Remark. This lemma is stronger than the one I originally proved and was suggested to me by Dana Scott. His proof was obtained by considering the Boolean algebra of sets of C mod finite sets. Proof. Let So, Si, be the sets of C. We will define P0, Px, by means of a sequence of infinite partitions (Po, (Pi, of N into infinite sets of C such that for every k and n with k<n, Pk belongs to <Pn and either Sk or S* is included in UJ<n Pj. Assume that these conditions hold for all n up through /. We shall define Pt and (Pt+x so that they also hold for n = t+l. Let S be S if S( has infinitely many infinite parts under (Pt; otherwise let S be S(. In either case, S has infinitely many infinite parts under <Pt. Let those infinite parts of S which are not contained in one of P0,, Pt~x be I0, ii,. Then put

4 1969] FINITELY GENERATED CLASSES 611 P, = 7o VJ (s - U P*\ Let di, a2, be the numbers which do not belong to UPiWU 7y, kit Jäl if there are any. Put T* = T* W {a*} if there is an ak; otherwise Tk = h- Finally, put <JVj = {Po,Pi,,Pt, Ti, }. Clearly PtQ<?t+i, SQÜknt Pk, and (P(+i is an infinite partition of TV into infinite sets of 6. Now let (P = { P0, Pi, }. By construction each Pk is an infinite set of 6, (P is disjoint, and for every k, either Sk or Sk is contained in U/s* Pj- Finally, every «belongs to some Py. In fact, {«} =S for some I, so w belongs to some P, with 7 not exceeding 1. Hence G> satisfies the lemma. Theorem 2. If a denumerable field of sets Q contains all finite sets and includes an infinite disjoint class (? of infinite sets Pa, Pi, -, then Q is finitely generated. Proof. Without loss of generality, we may suppose that (P is a partition of N; otherwise adjoin the missing numbers to the parts Pk with at most one to each P*. By Lemma 3, we may also assume that (? has the further property that for every S in C either S or S is included in a finite union of P's. By Lemma 2, we can define a function 77 so that 77(S) = P0 for every infinite set S in Q. Hence 77 is compatible with e. Let G be defined so that G (S) = P +i for every infinite set S of e which is included in Pn. The existence of G also follows from Lemma 2. Then GkH(N) = Pk for all k. Furthermore, if S is included in a finite union of P's, then G(S) is the union of a finite set and a finite union of P's; if S is included in a finite union of P's, then G(S) is the union of a finite set and the complement of a finite union of P's. In both cases G(S) belongs to Q. Hence G is compatible with C. We can now finish the proof by noticing the hypotheses of Theorem 1 hold. We can also construct F more directly as follows: Let S0, Si, be the nonempty sets of 6. By Lemma 2, there is a function F such that F(T) =S* if T is an infinite subset of P* which belongs to 6. Then F is compatible with C. Indeed, if S is contained in a finite union of P's then F(S) is the union of a finite number of sets of 6 and let

5 612 JULIA ROBINSON [June if S is contained in a finite union of P's then (S) =N. Every nonempty set in 6 is generated by, G, and H since Si = FGkH(N). Theorem 3. e/ Q be a denumerable field of sets which contains all finite sets. Suppose A0,, An is a partition of N into infinite sets of 6 none of which is the union of two disjoint infinite sets of Q. Then 6 is finitely generated if and only ifn>0. Proof. If n = 0 then Q is the field of all finite and cofinite sets. Suppose 6 were generated by i,, Fk, Gx,, G where the range of each is finite and the range of each G is cofinite. If G is obtained by composition from the G's then the range of G is cofinite. Hence if H is obtained by composition from the 's and G's then H has a finite range only if some, say F, is used to obtain H. Then the number of elements in H(N) is not more than the number of elements in Fj(N). Hence the number of elements in the finite sets generated by i,, Fk, Gx,, Gi is bounded which gives a contradiction. Thus for n = 0, 6 is not finitely generated. Suppose n>0. For KQ{0,, n}, let SK = {JkeK Ak. Every set of C differs from some Sk by a finite number of elements. We will first show that every SK can be generated by three functions A, B, and C which are compatible with 6. By Lemma 2, there is a function A such that A (S) = Ao for every infinite set Sin Q. A is compatible with C since A (S) is either finite or A0 for every S in 6. Let be a permutation which maps Ak onto Ak+x for k = 0, -, n 1 and maps An onto A0. B is compatible with 6 since B maps any set differing from SK by a finite set onto a set differing from SM by a finite set where M is obtained from K by adding 1 to each element of if other than n and replacing n by 0. Finally, let G be a function which is one-to-one on each Ai and maps Ai onto A0WA<. Again C is compatible with C since C maps any set differing from Sk by a finite set onto a set differing from S m by a finite set where M = 0 if K~ 0 and M=K\J{0} iík? 0. Suppose t is the least element of a nonempty set KQ {0,, n}. If K has just one element then SK B'A(N). If K has more than one element, let K! = {k:k + te K/\k> 0}. Then Sk = B'C(Sk>). Hence by induction on the number of elements of K, for every nonempty set K, there is a function H obtained from B and C by composition such that SK = HA(N). Now if S differs from SK (K nonempty) by a finite set, then there is a set E which differs from A0 by a finite set such that for some H

6 1969] FINITELY GENERATED CLASSES 613 obtained from B and C by composition H(E)=S. Indeed, we can choose 77as above so that H(A0) =SK; since 77(N) =N and H~1(j) is finite for all j, we can obtain E by modifying A0 by a finite set. Thus, it remains to show that there are a finite number of functions compatible with C which generate every finite set and every set differing from Ao by a finite set. Order the natural numbers in a sequence of the order type of the integers, say, a_i, a0, ait, so that A0 = {a0, c_i, a_2, } and for 7>0, ajqak if /=& (mod w) for k = l,, n. Now define three functions F, G, and 77 as follows: F(ak) = Ok+i, G(ak) = ak if k á 0, = ah+i if k > 0, 77(0*) = a0 if k Ú 0, = a* if 4 > 0. We will show that 6 is generated by A, B, C, F, F~\ G and 77. It is easy to check that F, F-1 G and 77 are all compatible with e. Now for j, k, I, greater than 0, and F(A0 U {a, ak, a, })= A0 U {ah a3-+i, ah+i, ai+i, } G(A0 W {a, ak, a, })= A0 U {ai+1, a*+i, a +i, }. Hence by repeatedly applying F and G, we can obtain a function which maps A0 onto the union of an arbitrary finite set and Ao. Next to obtain a set which differs from A0 by an arbitrary finite set, we apply a suitable power of F-1 to a set which is the union of a finite set and Ao. On the other hand, if we first apply 77 to the union of Ao and a finite set and then apply a suitable power of F or F"1, we can obtain an arbitrary nonempty finite set. Remark. This proof is due to R. M. Robinson. In the original proof, the number of generators depended on». Lemma A. If A can be partitioned into arbitrarily many infinite sets of C and A = B U C where B and C are infinite sets of C, then at least one of B and C can be partitioned into arbitrarily many infinite sets of Q. Proof. Let A be partitioned into 2«infinite sets Si,, S2n of e. Then B = {J(Br\Sk) and C = U(Cns ). Either B or C has at least «infinite parts under this partition of A and hence can be partitioned

7 614 JULIA ROBINSON into arbi- into n infinite sets of 6. Hence B or C can be partitioned trarily many infinite sets of Q. Theorem 4. If Q is a denumerable field of sets which contains all finite sets then either Q is the field of all finite and cofinite sets or 6 is finitely generated. Proof. Suppose there is a set A in C such that both A and A are infinite. If TV can be partitioned into at most n infinite sets of 6 for some n, then Theorem 3 applies. Otherwise N can be partitioned into arbitrarily many infinite sets of C and by Lemma 4 so can A or A. Now construct A0, Ai, in turn so that each Aj is an infinite set of C, and N (JjSk Aj can always be partitioned into arbitrarily many infinite sets of C. Of course, we must choose Aj to be disjoint from the earlier A's. Hence Theorem 2 applies and 6 is finitely generated. Reference 1. Julia Robinson, Finite generation of recursively enumerable sets, Proc. Amer. Math. Soc. 19 (1968), University of California, Berkeley